Cars II

Power Efficiency Guide

Ultimate Guide to Power Efficiency

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We estimated that a car driven 100 km uses about 80kWh of energy.

Where does this energy go? How does it depend on properties of the car? Could we make cars that are 100 times more efficient? Let's make a simple cartoon of car-driving, to describe where the energy goes. The energy in a typical fossil-fuel car goes to four main destinations, all of which we will explore:

1. speeding up then slowing down using the brakes;

2. air resistance;

Figure A.1. A Peugot 206 has a drag coefficient of 0.33. Photo by Christopher Batt.

3. rolling resistance;

4. heat - 75% of the energy is thrown away as heat, because the energy-conversion chain is inefficient.

Initially our cartoon will ignore rolling resistance; we'll add in this effect later in the chapter.

Assume the driver accelerates rapidly up to a cruising speed v, and maintains that speed for a distance d, which is the distance between traffic lights, stop signs, or congestion events. At this point, he slams on the brakes and turns all his kinetic energy into heat in the brakes. (This vehicle doesn't have fancy regenerative braking.) Once he's able to move again, he accelerates back up to his cruising speed, v. This acceleration gives the car kinetic energy; braking throws that kinetic energy away.

Energy goes not only into the brakes: while the car is moving, it makes air swirl around. A car leaves behind it a tube of swirling air, moving at a speed similar to v. Which of these two forms of energy is the bigger: kinetic energy of the swirling air, or heat in the brakes? Let's work it out.

The key formula for most of the calculations in this book is:

For example, a car of mass m = 1000 kg moving at 100 km per hour or v = 28m/s has an energy of mv2 ~ 390 000 J ~ 0.1 kWh.

Figure A.2. Our cartoon: a car moves at speed v between stops separated by a distance d.

The car speeds up and slows down once in each duration d/v. The rate at which energy pours into the brakes is:

kinetic energy time between braking events where mc is the mass of the car.

Figure A.3. A car moving at speed v creates behind it a tube of swirling air; the cross-sectional area of the tube is similar to the frontal area of the car, and the speed at which air in the tube swirls is roughly v.

The tube of air created in a time t has a volume Avt, where A is the cross-sectional area of the tube, which is similar to the area of the front view of the car. (For a streamlined car, A is usually a little smaller than the frontal area Acar, and the ratio of the tube's effective cross-sectional area to the car area is called the drag coefficient Cd-Throughout the following equations, A means the effective area of the car, Cd Acar- ) The tube has mass mair = pAvt (where p is the density of air) and swirls at speed v, so its kinetic energy is:

1 , 2 -pAvt vA, and the rate of generation of kinetic energy in swirling air is:

\pAvtv2

I'm using this formula:

mass = density x volume

The symbol p (Greek letter 'rho') denotes the density.

So the total rate of energy production by the car is:

power going into brakes + power going into swirling air

Both forms of energy dissipation scale as v3. So this cartoon predicts that a driver who halves his speed v makes his power consumption 8 times smaller. If he ends up driving the same total distance, his journey will take twice as long, but the total energy consumed by his journey will be four times smaller.

Which of the two forms of energy dissipation - brakes or air-swirling -is the bigger? It depends on the ratio of

If this ratio is much bigger than 1, then more power is going into brakes; if it is smaller, more power is going into swirling air. Rearranging this ratio, it is bigger than 1 if mc > pAd.

Now, Ad is the volume of the tube of air swept out from one stop sign to the next. And pAd is the mass of that tube of air. So we have a very simple situation: energy dissipation is dominated by kinetic-energy-being-dumped-into-the-brakes if the mass of the car is bigger than the mass of the tube of air from one stop sign to the next; and energy dissipation is dominated by making-air-swirl if the mass of the car is smaller (figure A.4).

Let's work out the special distance d* between stop signs, below which the dissipation is braking-dominated and above which it is air-swirling dominated (also known as drag-dominated). If the frontal area of the car is:

Figure A.4. To know whether energy consumption is braking-dominated or air-swirling-dominated, we compare the mass of the car with the mass of the tube of air between stop-signs.

Figure A.4. To know whether energy consumption is braking-dominated or air-swirling-dominated, we compare the mass of the car with the mass of the tube of air between stop-signs.

Figure A.5. Power consumed by a car is proportional to its cross-sectional area, during motorway driving, and to its mass, during town driving. Guess which gets better mileage - the VW on the left, or the spaceship?

t and the drag coefficient is Cd special distance is:

1000 kg

Pcd Ac

So "city-driving" is dominated by kinetic energy and braking if the distance between stops is less than 750 m. Under these conditions, it's a good idea, if you want to save energy:

1. to reduce the mass of your car;

2. to get a car with regenerative brakes (which roughly halve the energy lost in braking - see Chapter 20); and

3. to drive more slowly.

When the stops are significantly more than 750 m apart, energy dissipation is drag-dominated. Under these conditions, it doesn't much matter what your car weighs. Energy dissipation will be much the same whether the car contains one person or six. Energy dissipation can be reduced:

1. by reducing the car's drag coefficient;

2. by reducing its cross-sectional area; or

3. by driving more slowly.

The actual energy consumption of the car will be the energy dissipation in equation (A.2), cranked up by a factor related to the inefficiency of the engine and the transmission. Typical petrol engines are about 25% efficient, so of the chemical energy that a car guzzles, three quarters is wasted in making the car's engine and radiator hot, and just one quarter goes into "useful" energy:

total power of car ~ 4

p Av3

Let's check this theory of cars by plugging in plausible numbers for motorway driving. Let v = 70 miles per hour = 110km/h = 31m/s and A = CdAcar = 1 m2. The power consumed by the engine is estimated to be roughly

4 x ipAv3

If you drive the car at this speed for one hour every day, then you travel 110 km and use 80kWh of energy per day. If you drove at half this speed for two hours per day instead, you would travel the same distance and use up 20kWh of energy. This simple theory seems consistent with the

Energy-per-distance

Car at 110 km/h

Bicycle at 21 km/h ^

Planes at 900 km/h A380 27kWh/100 seat-km

Table A.6. Facts worth remembering: car energy consumption.

mileage figures for cars quoted in Chapter 3. Moreover, the theory gives insight into how the energy consumed by your car could be reduced. The theory has a couple of flaws which we'll explore in a moment.

Could we make a new car that consumes 100 times less energy and still goes at 70mph? No. Not if the car has the same shape. On the motorway at 70 mph, the energy is going mainly into making air swirl. Changing the materials the car is made from makes no difference to that. A miraculous improvement to the fossil-fuel engine could perhaps boost its efficiency from 25% to 50%, bringing the energy consumption of a fossil-fuelled car down to roughly 40 kWh per 100 km.

Electric vehicles have some wins: while the weight of the energy store, per useful kWh stored, is about 25 times bigger than that of petrol, the weight of an electric engine can be about 8 times smaller. And the energy-chain in an electric car is much more efficient: electric motors can be 90% efficient.

We'll come back to electric cars in more detail towards the end of this chapter.

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